Question

Find all eigenvalues and their corresponding eigenspaces of the following matrix. A.\( \begin{pmatrix}3&0\\ 1&2\end{pmatrix} \)

Answer 1

Solution

 A.\( \begin{pmatrix}3&0\\ 1&2\end{pmatrix} \) \( \)let \( \lambda \in \mathbb{R} \) be an eigenvalue of \( A \). Then

We have, \( Ax=\lambda x , \lambda \in \mathbb{R}^2, x=\begin{pmatrix}x_1\\ x_2\end{pmatrix}\in \mathbb{R}^2, x \neq 0 \)

\( \iff Ax-\lambda x=0 \)

\( \iff \bigg(A-\lambda I\bigg)x=0 \hspace{2mm}, \bigg(x \neq 0\bigg) \)

\( \implies |A-\lambda I|= \left|\begin{pmatrix}3&0\\ 1&2\end{pmatrix}-\lambda \begin{pmatrix}1&0\\ 0&1\end{pmatrix}\right|=\begin{vmatrix}3-\lambda &0\\ 1&2-\lambda \end{vmatrix}=0 \)

\( \implies \bigg(3-\lambda\bigg)\bigg(2-\lambda\bigg)=0 ,\implies \lambda_1=2 , \lambda_2=3 , \) is eigenvalue. Thus, \( specc(A)=\bigg({2,3}\bigg). \)

\( +For \hspace{3mm}\lambda_1=2 : \bigg(A-\lambda_1 x\bigg)=0 \hspace{3mm}, x=\begin{pmatrix}x_1\\ x_2\end{pmatrix}\in \mathbb{R}^2 \)

\( \iff \bigg(A-2I\bigg)x=0 \implies \begin{pmatrix}1&0\\ 1&0\end{pmatrix}x=0 \)

\( \implies \begin{pmatrix}1&0\\ 1&0\end{pmatrix}\:\sim \begin{pmatrix}1&0\\ 0&0\end{pmatrix},\hspace{2mm} Let, \hspace{2mm}x_2=t \implies x_1=0 \)

\( \implies x=\begin{pmatrix}x_1\\ x_2\end{pmatrix}=\begin{pmatrix}0\\ t\end{pmatrix} , t\in \mathbb{R}^2, \hspace{2mm} \)

x is a eigenvector corrsponding to \( \lambda_1 =2 \)

The eigenspace corrsponding to \( \lambda_2=2 \hspace{2mm}is\hspace{2mm} E_2=\left\{\begin{pmatrix}0\\ t\end{pmatrix}\in \mathbb{R}^2|t \in \mathbb{R}^*\right\} \)

\( +For \hspace{3mm} \lambda_2=3 : \bigg(A-\lambda_2I\bigg) x=0 \hspace{3mm}, x=\begin{pmatrix}x_1\\ x_2\end{pmatrix}\in \mathbb{({R}^*)}^2 \)

\( \iff \bigg(A-3I\bigg)x=0 \hspace{3mm} \implies \begin{pmatrix}0&0\\ 1&-1\end{pmatrix}x=0 \)

\( \begin{pmatrix}0&0\\ 1&-1\end{pmatrix}\sim \begin{pmatrix}1&-1\\ 0&0\end{pmatrix} \hspace{2mm},Let \hspace{2mm} x_1=t\implies x_2=t \)

\( \implies E_3=\left\{\begin{pmatrix}t\\ t\end{pmatrix}\in \mathbb{R}^2|t\in \mathbb{R^*}\right\}\hspace{2mm}Thus,\hspace{2mm} E_2=span\bigg(0,1\bigg) \implies dim(E_2)=1 \)

Answer :

Therefore  

\( specc(A)=\bigg({2,3}\bigg). \)

\( E_2=span\bigg(0,1\bigg) \implies dim(E_2)=1 \)